Integral over the manifold $S^1$

Question

Let $\omega=xdx$.

What is $\int_{S^1}\omega$?

What I know:
The final result should be $1$.
We can take parametrization $p:[0,\pi]\rightarrow S^1$ with $p(t)=(\cos(t),\sin(t))$ and $Dp(t)=(\sin(t),\cos(t))$. Then $$\int_{S^1}\omega=\int_{S^1}xdx=\int_{p([0,\pi]}xdx=\int_{[0,\pi]}p^*xdx=\int_{[0,\pi]}xdx(p(t)(D(p(t)).$$ Is this correct? How do I continue from here?

Answer 1

An interesting approach is to use Stokes, noting that $S^1$ is the boundary of the unit disk. Then $d\omega = dx\wedge dx=0$, so

$$\int_{S^1} \omega = \int_{B} d\omega=0$$

Alternatively, with the parametrization you gave the pullback of $\omega$ is

$$\cos t(\cos t)'=-\cos t\sin t=-\frac{\sin 2t}2$$

and this integrates to $0$ over $[0,2\pi)$.

Answer 2

Notice that the integral on $S^1$ and the integral on $S^1 \setminus \{(1,0)\}$ are equal, because $\{(1,0)\}$ is a null set. A convenient parametrization is $p : (0, 2\pi) \to S^1 \setminus \{(1,0)\}$ given by $p(t) = (\cos t, \sin t)$. Then, as you say,

$$\int \limits _{S^1} \omega = \int \limits _{S^1 \setminus \{(1,0)\}} \omega = \int \limits _{(0,2\pi)} p^* \omega = \int \limits _{(0,2\pi)} \cos t \ \Bbb d (\cos t) = \int \limits _0 ^{2\pi} \cos t (\cos t)' \ \Bbb d t = \frac {\cos^2 t} 2 \Bigg| _0 ^{2\pi} = \frac 1 2 - \frac 1 2 = 0 .$$